Understanding Gravity and Weight - High School Physics

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Question

Michael lands on a new planet. If his mass is and the acceleration due to gravity on this planet is $-1.56\frac{m}{s^2}$ , what would his weight be?

Answer

Weight is a particular force that is equal to mass times acceleration due to gravity. Start with Newton's second law, $\vec{F}=m\vec{a}$ .

$\vec{F}_{weight}=m\vec{a}_{gravity}$

Plug in the given values to solve.

$\vec{F}=(20kg)(-1.56\frac{m}{s^2})$

$\vec{F}_{weight}=-31.2N$

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Question

Leslie is an astronaut on a new planet. She knows her mass is , and she can calculate her weight on the planet to be . What is the acceleration due to gravity on the new planet?

Answer

Weight is a specific force. It is always equal to the mass times acceleration due to gravity: $\Vec{F}_{weight}=m\vec{a}_{gravity}$ .

Plug in the given values to solve for the acceleration.

$(-512N)=(70kg)(\vec{a})$

$\frac{-512N}{70kg}=\vec{a}$

$-7.31\frac{m}{s^2}=\vec{a}$

The gravitational acceleration is negative because it acts in the downward direction.

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Question

Jenny calculates that her weight on Earth is equal to . Assuming acceleration due to gravity on earth is equal to $-9.8\frac{m}{s^2}$ , what is her mass?

Answer

Weight is a specific force that is always equal to mass times acceleration due to gravity.

Start with Newton's second law: $\vec{F}=m\vec{a}$ .

$\vec{F}_{weight}=m\vec{a}_{gravity}$

Plug in the given values to solve for the mass.

$-500N=(m)(-9.8\frac{m}{s^2})$

$\frac{-500N}{-9.8\frac{m}{s^2}}=m$

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Question

Jerry wants to lift a ball with exactly enough force so that it's upward velocity is constant. How much force should he use?

$\small g=-9.8\frac{m}{s^2}$

Answer

If the velocity on an object is constant, that means it has no acceleration. If it has no acceleration, that means that the net force on the object is equal to zero. We can see this conclusion by using Newton's second law.

$F_{net}=ma$

$F_{net}=m(0\frac{m}{s^2})$

$F_{net}=0N$

Another way to think of $F_{net}$ is the sum of all the forces. Since the only two forces acting upon the ball are gravity and Jerry's lifting force, we can see: $F_{net}=F_{Jerry}+F_{gravity}$ .

Since the net force is zero, the magnitude of Jerry's force must equal the magnitude of the force of gravity, but in the opposite direction.

$0=F_{Jerry}+F_{gravity}$

$-F_{Jerry}=F_{gravity}$

This means that once we find $F_{gravity}$ , then Jerry's lifting force will be the same magnitude but in the opposite direction. Use Newton's second law to find the force of gravity.

$F_{gravity}=ma$

$F_{gravity}=(0.12kg)(-9.8\frac{m}{s^2})$

$F_{gravity}=-1.18N$

This means that Jerry's lifting force will be .

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Question

The acceleration of gravity on the moon is significantly less than the acceleration of gravity on earth. What will happen to an astronaut's weight and mass on the moon, compared to her weight and mass on Earth?

Answer

Mass is a measure of how much matter is in an object, while weight is a measurement of the effective force of gravity on the object.

The amount of matter in the astronaut does not change; therefore, she has the same mass on the moon as she has on Earth.

Her weight, however, will change due to the change in gravity. The force of gravity will be the product of the acceleration and the astronaut's mass.

If the gravity is less on the moon than on Earth, then the force of gravity on the astronaut will also be less on the moon; thus, she will weigh less.

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Question

If the mass of the object is and $\theta=13^\circ$ , what is the value of W?

$\small g=-9.8\frac{m}{s^2}$

Answer

W will be the weight of the object. Weight is a very specific force: it is the mass times gravity. As it turns out, the angle is irrelevant in finding weight.

Using Newton's second law and the given values for mass and gravity, we can solve for W.

$W=(18.5kg)*(-9.8\frac{m}{s^2})$

Note that the weight is negative, because it is acting in the downward direction.

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Question

An astronaut lands on a new planet and discovers her weight on this planet is half of her weight on Earth. What is the acceleration due to gravity on this planet in terms of the acceleration due to gravity on Earth ( $\small g$ )?

Answer

Weight is a very specific force, equal to the mass times gravity:

On this new planet her weight is half of what it was on Earth. We can write this mathematically, using the weight equation.

$F_{W2}=\frac{1}{2}F_{W1}$

$mg_{new}=\frac{1}{2}mg_{Earth}$

Mass cancels out from each side, leaving a relationship between the gravitational accelerations.

$g_{new}= \frac{1}{2}g_{Earth}$

That means the acceleration due to gravity on this new planet is half of what it was on earth.

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Question

A ball that weighs on Earth weighs on a recently discovered planet. What is the force of gravity on this new planet?

Give your answer with the correct number of significant figures.

Answer

Weight is defined as the force of gravity on an object. We can use Newton's second law to write an equation for weight.

If the ball weighs on Earth, then its mass can be found using this equation and the acceleration of gravity on Earth.

$4N = m(9.8\frac{m}{s^2})$

$m = \frac{4N}{(9.8\frac{m}{s^2})} = 0.40816 kg$

Use this mass and the given weight on the new planet to find the acceleration of gravity on this new planet. Though our initial values (and thus our final values) only allow one significant figure, we will not round until the end of all calculations. This ensures that we preserve accuracy before adjusting for precision.

$\frac{19N}{0.40816kg} = 46.55\frac{m}{s^2}$

Adjust this value to one significant figure by rounding up. The zero in the tens place is before the decimal, and is not considered significant.

$46.55\frac{m}{s^2}\approx 50\frac{m}{s^2}$

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Question

A satellite is in orbit above the Earth. What is the relationship between the acceleration due to gravity on the satellite, versus the acceleration due to gravity on the Earth's surface?

Answer

For this problem, we are comparing the force of gravity on the surface, or weight, to the force of gravity on the satellite. We can use Newton's second law to find the weight of the satellite, and the law of universal gravitation to find the gravity on the satellite. These two terms will be equal to one another.

$F_G=G\frac{m_1m_2}{r^2}$

$mg=G\frac{m_1m_2}{r^2}$

Let's call the Earth and the mass of the satellite.

$m_sg=G\frac{m_Em_s}{r^2}$

Notice that the masses of the satellite cancel out.

$g=G\frac{m_E}{r^2}$

This formula gives us the acceleration of gravity in terms of the mass of the Earth and the distance from the Earth's center. We can write two separate equations, one for the surface and one for the satellite. Since the mass of the Earth doesn't change and is a constant, the only variable that can change is , the distance between the objects.

$g_{surface}=G\frac{m_E}{r_E^2}$

$g_{satellite}=G\frac{m_E}{(r_E+h)^2}$

On Earth, is the radius of the earth. For the satellite, is the radius of the Earth PLUS the orbiting distance; therefore . Because we are dividing by our , a greater gives us a smaller .

The satellite in space will have a smaller acceleration due to gravity. It will not be zero, but it will be smaller than the acceleration on the surface.

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Question

A satellite orbits above the Earth. What is the satellite's acceleration due to gravity?

$m_E=5.9710^{24}kg$

$G=6.6710^{-11}\frac{m^3}{kg*s}$

Answer

To solve this problem, use the law of universal gravitation.

$F_G=G\frac{m_1m_2}{r^2}$

Remember that is the distance between the centers of the two objects. That means it will be equal to the radius of the earth PLUS the orbiting distance.

Use the given values for the masses of the objects and distance to solve for the force of gravity.

$F_G=G\frac{m_sm_E}{(r_E+h)^2}$

$F_G=(6.6710^{-11}\frac{m^3}{kgs})\frac{(2500kg)(5.9710^{24}kg)}{(6.3710^5m+3.810^8m)^2}$

$F_G=(6.6710^{-11}\frac{m^3}{kgs^2})\frac{1.4910^{28}kg^2}{1.4510^{17}m^2}$

$F_G=(6.6710^{-11}\frac{m^3}{kgs^2})(1.0310^{11}\frac{kg^2}{m^2})$

Now that we know the force, we can find the acceleretion. Remember that weight is equal to the mass times acceleration due to gravity.

Set our two forces equal and solve for the acceleration.

$2.710^{-3}\frac{m}{s^2}=g$

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Question

A satellite orbits above the Earth. What is the gravitational acceleration on the Earth caused by the satellite?

$m_E=5.9710^{24}kg$

$G=6.6710^{-11}\frac{m^3}{kg*s}$

Answer

To solve this problem, use the law of universal gravitation.

$F_G=G\frac{m_1m_2}{r^2}$

Remember that is the distance between the centers of the two objects. That means it will be equal to the radius of the earth PLUS the orbiting distance.

Use the given values for the masses of the objects and distance to solve for the force of gravity.

$F_G=G\frac{m_sm_E}{(r_E+h)^2}$

$F_G=(6.6710^{-11}\frac{m^3}{kgs})\frac{(2500kg)(5.9710^{24}kg)}{(6.3710^5m+3.810^8m)^2}$

$F_G=(6.6710^{-11}\frac{m^3}{kgs^2})\frac{1.4910^{28}kg^2}{1.4510^{17}m^2}$

$F_G=(6.6710^{-11}\frac{m^3}{kgs^2})(1.0310^{11}\frac{kg^2}{m^2})$

Now that we know the force, we can find the acceleretion. Remember that weight is equal to the mass times acceleration due to gravity.

.

Set our two forces equal and solve for the acceleration.

$6.87N=5.9710^{24}kgg$

$\frac{6.87N}{5.9710^{24}kg}=g$

$1.15*10^{-24}\frac{m}{s^2}=g$

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Question

An object is placed in the direct center of the Earth. What would be the perceived weight of the object?

Answer

We must use Newton's law of universal gravitation to solve this question.

$F_G=G\frac{m_1m_2}{r^2}$

There are three variables that really change the force of gravity: the mass of each object and the distance between the bodies.

The important thing to recognize here, though, is that when an object is in the center of the earth, the mass of the earth is distributed symmetrically all around it. It's like being in the center of a giant bubble. Because the mass is symmetrically distributed, the mass that is trying to pull the object in each direction is equal. Essentially, the mass pulling upward cancels out the mass pulling downward, and the mass pulling right cancels out the mass pulling left.

This happens for the entirety of the circle, leaving you with a net force of zero acting upon the object.

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Question

A woman stands on the edge of a cliff and drops two rocks, one of mass and one of , from the same height. Which one experiences the greater force?

Answer

The formula for force is given by Newton's second law:

Both rocks will experience the same acceleration, , or the acceleration due to gravity.

Use the mass of each rock in this equation to find which rock experiences a greater force.

$F_{g,m}=(m)g$

$F_{g,3m}=(3m)g=3(mg)$

We can see that the force on the rock with mass of is equal to three times for the force on the rock with mass of . The heavier rock experiences the greater force.

$F_{g,3m}=3(F_{g,m})$

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Question

A woman stands on the edge of a cliff and drops two rocks, one of mass and one of , from the same height. Which one experiences the greater acceleration?

Answer

Even though the rocks have different masses, the acceleration on both will be , the acceleration due to gravity. We can look at Newton's second law to see the force experienced by the rocks:

When objects are in free-fall, the acceleration will be equal to the acceleration from gravity, regardless of the mass of the object.

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Question

A ball falls off a cliff. What is the force of gravity on the ball?

$g=-9.8\frac{m}{s^2}$

Answer

Newton's second law states:

In this case the acceleration will be the constant acceleration due to gravity on Earth.

Use the acceleration of gravity and the mass of the ball to solve for the force on the ball.

$F=(0.05kg)(-9.8\frac{m}{s^2})$

The answer is negative because the force is directed downward. Since gravity is always acting downward, a force due to gravity will always be negative.

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Question

An astronaut weighs on Earth. On a distant moon, she weighs . What is the acceleration due to gravity on this moon?

$g=-9.8\frac{m}{s^2}$

Answer

First we need to find the mass of the astronaut using Newton's second law.

We know the total weight of the astronaut and the acceleration due to gravity on Earth, allowing us to solve for her mass.

$150N=m*9.8\frac{m}{s^2}$

$\frac{150N}{9.8\frac{m}{s^2}}=m$

Now that we know her mass, we can look at her weight on the distant moon. We know her weight and mass, allowing us to solve for the acceleration due to gravity in this new environment.

$\frac{130N}{15.3kg}=a$

$8.5\frac{m}{s^2}=a$

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Question

The mass of the moon is less than that of Earth, causing it to have a gravitational acceleration less than $9.8\frac{m}{s^2}$ . Which of the following could be the weight of an object on the moon, if the object weighs on Earth?

Answer

Newton's second law states that:

We know from the problem that the acceleration due to gravity on the moon is less than the acceleration due to gravity on Earth. The mass of the object, however, will remain constant. The result is that the force of gravity on the object while on the moon will be less than the force on the object while on Earth.

$F_{g\ moon}<F_{g\ Earth}$

This means that the weight of the object while on the moon must be less than . Since the object has a weight on Earth, however, we know that its weight on the moon cannot be zero. This would imply that either the acceleration due to gravity on the moon is zero, or that the mass is zero, neither of which is possible. This allows us to eliminate from the answers.

The only other option that is less than is .

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Question

An astronaut has a mass of and Mars has an acceleration due to gravity of $3.8\frac{m}{s^2}$ . What is her weight on Mars?

Answer

Weight is a very specific force, determined by the acceleration due to gravity acting on a given mass. Using Newton's second law, we can see that weight will be equal to the equation:

We are given the mass of the astronaut and the acceleration due to gravity on Mars. Using these values, we can calculate her weight on Mars.

$F_g=(65kg)(3.8\frac{m}{s^2})$

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